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LME49724 Input Resistance

Mark Hiteshew
Prodigy 70 points

Other Parts Discussed in Thread: LME49724

I am trying to determine the Input Resistance of my op-amp circuit to see if I am loading my source down. I have the op-amp set up like the demo board on p. 22 of the data sheet for this opamp (http://www.ti.com/lit/ds/symlink/lme49724.pdf). I am inputting a single ended signal, so the negative input is grounded and the positive input is my signal. I want to know how to calculate the input resistance for my signal. Because of the differential output, I cannot assume that the signal is simply going through the resistor (R1) directly to ground. Any suggestions?

Thanks!

Mark

  • 0
    TI__Expert 3840 points

    Hi Mark,

    What is the impedance of your source?

    If we assume a non-inverting amplifier topology, from  Figure 16 pg 22, (R2//R4) = 500 ohms << rd the differential input impedance.

    The input resistance, Ri = rd(1 + ABeta) where Beta = R2/(R2+R4). If we assume for grins that rd = 1Mohm and A is 125dB or (1.8E6) and Beta is 0.5, then Ri is 9E11.

    If you think your source is loading down the input you could try increasing your gain setting resistor values or add a buffer.

    Best Regards,

    JD

  • 0 in reply to John De Celles
    Prodigy 70 points

    JD,

    Where did that equation come from? from this application note (http://www.analog.com/static/imported-files/application_notes/AN-1026.pdf) p. 3 on the right hand side, they determine that the input resistance is essentially 2*Rg where Rg is the value of the two input resistors.

    Mark

  • 0 in reply to Mark Hiteshew
    TI__Expert 3840 points

    Hi Mark,

    I thought that you were asking about the input impedance of the amplifier loading down your circuit, hence the reference to the differential input impedance, open loop gain and beta. I did reference the gain setting resistors implying that they would be the true impedance loading down your source.

    I asked about your source impedance because most sources are typically quite low such that your gain setting resistors wouldn't load it down. If the source impedance is higher you can increase your gain resistors as a check for validation of the cause.

    The AD app note is correct; RINdm = 2xRG.

    I'm sorry for the confusion.

    Best Regards,

    JD

  • 0 in reply to John De Celles
    Prodigy 85 points

    Same topic - input inpedance of the LME49724.

    So, if I wanted to design for a 600 Ohm input impedance  I would have the series input resistors R1 and R2 (from page 22 Figure.16) as 300 Ohm resistors?  

    And then the feedback resistors R3 and R4 would just be sized in order to give me the amount of gain I want? 

    Thanks,
    Eric 

  • 0 in reply to Eric Koester
    Prodigy 70 points

    Yes, Eric, you are correct

    Mark